# Graph Vout vs Vin of diode circuit

1. Mar 10, 2013

### leroyjenkens

1. The problem statement, all variables and given/known data
[PLAIN]http://postimage.org/image/bd4pyoqkt/ [Broken]

Make a sketch of Vout vs Vin for Vin from 0 to 5 V, assuming that both diodes have a diode drop of 0.5 V

2. Relevant equations

None that I know of.

3. The attempt at a solution

The way I see it, the Vout will start at 2V, since that's above the diode drop voltage, and that's what the middle battery is providing.
But as I increase Vin, it will just keep increasing Vout as the current flows directly to Vout. Why would it take the other path with the diodes?

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2. Mar 10, 2013

### CWatters

No that's not correct. What happens to the left diode when Vout > 2V + 0.5V

What happens to the right hand diode when Vout < 2V - 0.5V (eg when Vin is 0V).

3. Mar 10, 2013

### leroyjenkens

Is there an equation I can look at to figure this out? This stuff makes zero sense to me, and the book is zero help for these questions.

4. Mar 10, 2013

### rude man

Sure.
For a diode, i = 0, V < 0.5V
V = 0.5V, i > 0.

5. Mar 11, 2013

### CWatters

Ok this is how I would look at the circuit.

You can replaced Vin by a short circuit because Vin is an ideal voltage source (zero resistance) and ideal voltage sources can also sink current.

Since Vin = 0 is < 2V there is the possibility current may flow from the 2V source to 0V. Having established that you then look at what's in that path to confirm current will actually flow that way... In that path there is the right hand diode and a resistor in series.

So I can redraw the circuit as shown below (version on the left). I have removed the left hand diode because it is reverse biased.

You can then redraw the circuit to give that shown on the right hand side. 2V-0V= 1.5V which is greater than the 0.5V Vf of the diode. So the diode is forward biased and drops 0.5V so vout = 2V-0.5V = 1.5V

So we can draw a point on the graph at Vin=0, Vout = 1.5V.

More in a moment.

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6. Mar 11, 2013

### CWatters

Next I would look back at your original circuit.

If Vout was exactly 2V then no current would flow through either diode because both ends are at the same voltage. Therefore you can remove both diodes and you get this circuit below.

No current flows through the resistor so Vout = Vin = 2V.

You can draw a point on the graph at Vin = 2V and Vout = 2V.

If you think about it neither diode will conduct until Vf > 0.5V so Vout will equal Vin over the range Vout = 2V +/- 0.5V. (eg from 1.5V to 2.5V).

So you can add a straight line to the graph showing Vout=Vin over that range.

More in a moment..

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7. Mar 11, 2013

### CWatters

If Vin were increased to say 5V then that is greater than 2V so there is the possibility current may flow from the 5V source to the 2V source. Having established that you then look at what's in that path to confirm current will actually flow that way... In that path there is the left hand diode and a resistor in series. The right hand diode is reverse biased and can be removed.

The circuit therefore becomes that shown below...

5V-2V is more than the 0.5V Vf of the diode so the diode is conducting and will drop 0.5V. Therefore Vout cannot go more than 0.5V above the 2V source. eg Vout cannot go over 2.5V regardless of Vin.

So you can draw a point at Vin = 5V, Vout = 2.5V.

In fact you can draw a horizontal line from the point Vin = 5V, Vout = 2.5V towards the left until it meets the Vin=Vout line you drew in step two 2.

Likewise you can also draw a horizontal line from the point you drew in step one (Vin=0, Vout =1.5) to the right until until it meets the Vin=Vout line you drew in step two 2.

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8. Mar 11, 2013

### CWatters

PS. The circuit is called a clipper. Imagine Vin was an irregular wave form that looked like a mountain range. Vout would look similar but the peaks above 2.5V and the valleys below 1.5V will be "clipped off".

I might get a slapped wrist for giving too much help but you did say this stuff makes zero sense.

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